Question

A 2.0 kg stone is tied to a 0.50 m string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is:_____a. 0b. 6.0Nc. 8Nd. 10N

Answer 1

Answer:

The magnitude of the torque about the origin is 108 N-m.

Explanation:

Given that,

Mass of the stone, m = 2 kg

Length of the string, l = 0.5 m

Angular velocity of the stone, $\omega=12\ rad/s$

The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin, r = 0.75 m

Firstly it is required to find the centripetal force acting on the stone. It is given by :

$F=mr\omega^2$

$F=2\times 0.5\times (12)^2$

F = 144 N

Now the magnitude of the torque about the origin is given by :

$\tau=r\times F$

$\tau=0.75\ m\times 144\ N$

$\tau=108\ N-m$

So, the magnitude of the torque about the origin is 108 N-m. Hence, this is the required solution.